To evaluate the integral [tex]\(\int \frac{2x}{x^2 + 1} \, dx\)[/tex], we can follow these steps:
1. Identify the integrand:
The integrand is [tex]\(\frac{2x}{x^2 + 1}\)[/tex].
2. Consider a substitution:
Notice that the derivative of the denominator [tex]\(x^2 + 1\)[/tex] is [tex]\(2x\)[/tex], which appears in the numerator. Let’s use the substitution method:
[tex]\[
u = x^2 + 1
\][/tex]
Then, differentiate both sides with respect to [tex]\(x\)[/tex]:
[tex]\[
\frac{d u}{d x} = 2x \implies du = 2x \, dx
\][/tex]
Thus, the integral can be rewritten in terms of [tex]\(u\)[/tex]:
[tex]\[
\int \frac{2x}{x^2 + 1} \, dx = \int \frac{1}{u} \, du
\][/tex]
3. Integrate with respect to [tex]\(u\)[/tex]:
The integral [tex]\(\int \frac{1}{u} \, du\)[/tex] is a standard integral, which is known to be:
[tex]\[
\int \frac{1}{u} \, du = \ln |u| + C
\][/tex]
4. Substitute back [tex]\(u = x^2 + 1\)[/tex]:
Replace [tex]\(u\)[/tex] with [tex]\(x^2 + 1\)[/tex] to get the final answer in terms of [tex]\(x\)[/tex]:
[tex]\[
\int \frac{2x}{x^2 + 1} \, dx = \ln (x^2 + 1) + C
\][/tex]
Therefore, the solution to the integral [tex]\(\int \frac{2x}{x^2 + 1} \, dx\)[/tex] is:
[tex]\[
\boxed{\log(x^2 + 1) + C}
\][/tex]
where [tex]\(C\)[/tex] is the constant of integration.
